Question

For what value of k are the points (8 1) (3 -2k ) (k -5) colliner

Answer 1

Answer:

There are two values of k for which the points are collinear:

k = 2

and

k = 11 / 2

Step-by-step explanation:

The slope of the tangent from (8, 1) to (3, -2k) is

( -2k - 1 ) / ( 3 - 8 ) = ( 2k + 1 ) / 5

The slope of the tangent from (8, 1) to (k, -5) is

( -5 - 1 ) / ( k - 8 ) = -6 / ( k - 8 ).

The three points are collinear

<=> the two lines through (8,1) are the same line

<=> the gradients are equal

<=> ( 2k + 1 ) / 5 = -6 / ( k - 8 )

<=> ( k - 8 ) ( 2k + 1 ) = -30

<=> 2k² - 15k + 22 = 0

<=> ( k - 2 ) ( 2k - 11 ) = 0

<=> k = 2 or k = 11 / 2